How can I count only NULL values in Oracle/PLSQL?
I want to count only the null values. Is there a function that does that?
I don't know Oracle specifally, but ANSI SQL, COUNT(rowName) does not count NULL values, but COUNT(*) does. So you can write
SELECT COUNT(*) FROM YourTable WHERE YourColumn IS NULL
which counts the rows in YourTable that have YourColumn set to NULL.
As an alternative to mdma's response.
If you don't want to put a filter in the where you can
SELECT COUNT(case when xxx IS NULL THEN 1 end) cnt_xxx_null
FROM table
The Oracle documentation states that:
All aggregate functions except
COUNT(*) and GROUPING ignore nulls.
You can use the NVL function in the
argument to an aggregate function to
substitute a value for a null.
As an example, using the scott schema:
SQL> select empno, sal, comm
2 from emp;
EMPNO SAL COMM
---------- ---------- ----------
7369 800
7499 1600 300
7521 1250 500
7566 2975
7654 1250 1400
7698 2850
7782 2450
7788 3000
7839 5000
7844 1500 0
7876 1100
7900 950
7902 3000
7934 1300
14 rows selected.
You can see that the Comm column has 4 known values (i.e. Not null) and 10 unknown values (i.e. Null)
As count(your_column_name) ignores nulls you need to substitute the unknown values for something you can refer to. This can be achieved using the NVL function.
SQL> select count(nvl(comm, -1)) "number of null values"
2 from emp
3 where nvl(comm, -1) = -1;
number of null values
---------------------
10
I have used the value "-1" as the "alias" for my null values because I know that "-1" is not an existing value within the comm column.
EDIT:
Following Rob's suggestion. It is possible to remove the where clause from the above example and use the NVL2 function as shown below:
SQL> select count(nvl2(comm,null,-1)) "number of null values"
2 from emp
3 /
number of null values
---------------------
10
If you wants to count other values too with null then use of COALESCE function will improves execution time
Oracle Differences between NVL and Coalesce
SELECT COUNT(COALESCE( _COLUMN, 1)) AS CNT FROM _TABLE
I might try to inverse the null, see results
SELECT
COUNT(DECODE(YourField, null, 1, null)) Nulls,
count(*) Everything,
COUNT(YourField) NotNulls
FROM YourTable
Everything should equal nulls + notnulls
select count(nvl(values, 0)) from emp where values is null;
Function:
create or replace function xxhrs_fb_count_null
return number
as
l_count_null number;
begin
select count(*) into l_count_null from emp where comm is null;
return l_count_null;
end;
Usage:
select xxhrs_fb_count_null from dual
I believe your requirement is as below:
Table emp has 100 rows. Against 20 employees, HIRE_DATE column is NULL. So basically, you want to get 20 as output.
This is another method apart from the answers given by other contributors in this forum.
-- COUNT (1) would return 100
-- COUNT (hire_date) would return 80
-- 100 - 80 = 20
SELECT COUNT (1) -
COUNT (hire_date)
AS null_count
FROM emp;
Related
I have an issue while opening the text file of database output. The null value columns are replaced with the columns which have data. So the data is not properly formatting in the text file. I generated the text file from oracle sql developer. And after opening the text file I can see the alignment of columns is not proper(misplacing of values between null columns and columns which have data ). Can anyone please help me to solve this issue
Just guessing here, but - what you said reminds me of this. Example is based on Scott's EMP table where some employees have NULL in the COMM column:
SQL> select ename, job, comm, deptno
2 from emp
3 where deptno = 30;
ENAME JOB COMM DEPTNO
---------- --------- ---------- ----------
ALLEN SALESMAN 300 30
WARD SALESMAN 500 30
MARTIN SALESMAN 1400 30
BLAKE MANAGER 30
TURNER SALESMAN 30
JAMES CLERK 30
6 rows selected.
If you're spooling such data by concatenating columns, you get mess which is difficult to read:
SQL> select ename||'-'||job||'-'||comm||'-'||deptno
2 from emp
3 where deptno = 30;
ENAME||'-'||JOB||'-'||COMM||'-'||DEPTNO
----------------------------------------------------------
ALLEN-SALESMAN-300-30
WARD-SALESMAN-500-30
MARTIN-SALESMAN-1400-30
BLAKE-MANAGER--30
TURNER-SALESMAN--30
JAMES-CLERK--30
6 rows selected.
But, if you right-pad strings to certain length (RPAD function) and use NVL for missing data, then the result is somewhat prettier:
SQL> select rpad(ename, 10, ' ') ||'-'||
2 rpad(job , 10, ' ') ||'-'||
3 nvl(to_char(comm, '9990D00'), ' ') ||'-'||
4 deptno as result
5 from emp
6 where deptno = 30;
RESULT
---------------------------------------------------------------------
ALLEN -SALESMAN - 300,00-30
WARD -SALESMAN - 500,00-30
MARTIN -SALESMAN - 1400,00-30
BLAKE -MANAGER - -30
TURNER -SALESMAN - -30
JAMES -CLERK - -30
6 rows selected.
SQL>
I need to modify my script using rowcount to check if the data in table or not?. Here, i write the query to select a data for last 5 days from current system date. But sometimes there is no data in table for 5 days. So i need to fetch for 10 day or more.
Query:
Select ep.ENTERPRISE_NAME||'|'||s.id||'|'||s.SUBMISSION_DATE||'|'||E.VALUE
from JOB_SUMMARY_EXT e, ob_summary s, enterprise ep
where e.id = s.id and e.name_res_key = 'Model'
and s.job_id in (select id from job_summary where
trunc(start_date) > trunc(sysdate) -10 and service_name ='Model2' )
I don't know how to modify my Query using rowcount. If rowcount is 0 then i want select data for 10 days.Otherwise it should to fetch for 5 days automatically. I want this to be done as single query.
It looks that you want to select the last 5 "days" from that table. So, why would you anchor to SYSDATE if there aren't rows for each of those days? I'd suggest another approach: literally, select last 5 days. Here's how.
As I don't have your tables, I'm using Scott's EMP table which contains information about employees. It is an ancient one so HIREDATE column is set to 1980s, but never mind that. Sorting employees by HIREDATE in descending order shows:
SQL> alter session set nls_date_format = 'dd.mm.yyyy';
Session altered.
SQL> select ename, hiredate from emp order by hiredate desc;
ENAME HIREDATE
---------- ----------
ADAMS 12.01.1983 1.
SCOTT 09.12.1982 2.
MILLER 23.01.1982 3.
FORD 03.12.1981 4.
JAMES 03.12.1981 4.
KING 17.11.1981 5. --> I want to fetch rows up to KING
MARTIN 28.09.1981
TURNER 08.09.1981
CLARK 09.06.1981
BLAKE 01.05.1981
JONES 02.04.1981
WARD 22.02.1981
ALLEN 20.02.1981
SMITH 17.12.1980
14 rows selected.
SQL>
As you can see, the 4th date is shared by two employees so I want to include them both. DENSE_RANK analytic function helps:
SQL> with last5 as
2 (select ename,
3 job,
4 sal,
5 hiredate,
6 dense_rank() over (order by hiredate desc) rnk
7 from emp
8 )
9 select ename, job, sal, hiredate
10 from last5
11 where rnk <= 5;
ENAME JOB SAL HIREDATE
---------- --------- ---------- ----------
ADAMS CLERK 1100 12.01.1983
SCOTT ANALYST 3000 09.12.1982
MILLER CLERK 1300 23.01.1982
JAMES CLERK 950 03.12.1981
FORD ANALYST 3000 03.12.1981
KING PRESIDENT 5000 17.11.1981
6 rows selected.
SQL>
What does it do? The LAST5 CTE sorts employees (as above), DENSE_RANK ranks them; finally, the last SELECT (which begins at line #9) fetches desired rows.
In your case, that might look like this:
with last5 as
(select id,
dense_rank() over (order by start_date desc) rnk
from job_summary
where service_name = 'Model2'
)
select ep.enterprise_name,
s.id,
s.submission_date,
e.value
from job_summary_ext e
join ob_summary s on e.id = s.id
join last5 t on t.id = s.id
join enterprise ep on <you're missing join condition for this table>
where e.name_res_key = 'Model';
Note that you're missing join condition for the ENTERPRISE table; if that's really so, no problem - you'd use cross join for that table, but I somehow doubt that you want that.
Finally, as you use SQL*Plus, perhaps you don't need to concatenate all columns and separate them by the pipe | sign - set it as a column separator, e.g.
SQL> set colsep '|'
SQL>
SQL> select deptno, dname, loc from dept;
DEPTNO|DNAME |LOC
----------|--------------|-------------
10|ACCOUNTING |NEW YORK
20|RESEARCH |DALLAS
30|SALES |CHICAGO
40|OPERATIONS |BOSTON
SQL>
If you want to
return 10 last days if select count(*) returns 0, or
return 5 last days if select count(*) returns a positive number
then something like this might help (again based on Scott's EMP table):
with
tcnt as
-- count number of rows; use your own requirement, I'm checking
-- whether someone got hired today. In Scott's EMP table, nobody was
-- so CNT = 0
(select count(*) cnt
from emp
where hiredate >= trunc(sysdate)
)
select e.ename, e.job, e.sal, e.hiredate
from emp e cross join tcnt c
where e.hiredate >= case when c.cnt = 0 then trunc(sysdate) - 10
else trunc(sysdate) - 5
end;
Apply it to your tables; I don't know which of those 3 tables' count you want to check.
Tried to add in comments but it was too long for comments and Not clear on count based on but here is case in where clause substitute your count statement with nvl function
SELECT ep.ENTERPRISE_NAME||'|'||s.id||'|'||s.SUBMISSION_DATE||'|'||E.VALUE
FROM JOB_SUMMARY_EXT e,
ob_summary s,
enterprise ep
WHERE e.id = s.id
AND e.name_res_key = 'Model'
AND s.job_id IN
(SELECT id
FROM job_summary
WHERE service='Model'
AND trunc(start_date) >
CASE WHEN
(WRITE your SELECT COUNT criteria WITH NVL FUNCTION)<=0 THEN
trunc(sysdate) -10
ELSE trunc(sysdate)-5
END )
I find in PROD procedure code like:
select *
from (select "EMPNO",
"SAL",
"COMM",
"DEPTNO",
DENSE_RANK() OVER (PARTITION BY deptno ORDER BY null) AS drank
from SCOTT."EMP"
where deptno in (10,30))
where drank = 1
order by deptno
RESULT:
EMPNO SAL COMM DEPTNO DRANK
7934 1300 - 10 1
7839 5000 - 10 1
7782 2450 - 10 1
7844 1500 0 30 1
7900 950 - 30 1
7654 1250 1400 30 1
7499 1600 300 30 1
7698 2850 - 30 1
7521 1250 500 30 1
As result drank is always equal to 1. This is also true for:
DENSE_RANK() OVER (ORDER BY null) AS drank
DENSE_RANK() OVER (PARTITION BY comm ORDER BY null) AS drank
DENSE_RANK() OVER (PARTITION BY 1 ORDER BY null) AS drank
DENSE_RANK() OVER (PARTITION BY null ORDER BY null) AS drank
Is there any case when drank is not equal to 1 when there is ORDER BY null clause?
EDIT: I know dense_rank start with 1. Question is about values greater than 1.
The documentation says:
Use the order_by_clause to specify how data is ordered within a partition. For all analytic functions you can order the values in a partition on multiple keys, each defined by a value_expr and each qualified by an ordering sequence.
Within each function, you can specify multiple ordering expressions. Doing so is especially useful when using functions that rank values, because the second expression can resolve ties between identical values for the first expression.
Whenever the order_by_clause results in identical values for multiple rows, the function behaves as follows:
CUME_DIST, DENSE_RANK, NTILE, PERCENT_RANK, and RANK return the same result for each of the rows.
...
When you order by null, the order_by_clause results in identical values for multiple all rows (in the partition), so they all get the same result.
The documentation for dense_rank also says:
The ranks are consecutive integers beginning with 1.
So they get the same result, which has to be 1.
I have two tables in my database they are EXP1 and EXP2. I tried with the below query, this query is working when both the tables have same number of columns but my table EXP1 has 1000 columns n EXP2 has 1000+4.
select *
from
(
(select * from exp1
minus
select * from exp2)
union all
(select * from exp2
minus
select * from exp1)
);
INTRO: Below I show how one can do "by hand" what the tools (SQL Developer for example) can do much faster and much better. My interest in this (and yours!) is two-fold: learn and use some ideas that can help in many other problems; and understand what those tools do under the hood in the first place.
OK. Suppose you have two tables, and they have many columns in common (possibly not in the same order) and a few columns may be different - there may be a handful of columns in one table but not in the other. First you want to be able to look just at the common columns.
Then, suppose that's done. Now what's left of the two tables has many rows in common, but there are a few that are different. A row may exist in one table but not in the other, or two rows, one from each table, may be very similar but they may differ in just one or a small number of column values. Logically these are still one row in the first table but not the second, and the other row only in the second table but not in the first. However, let's say both tables have the same PK column - then you may have the same PK value in both tables, but at least one of the OTHER columns has different values for that PK value in the two tables. And, you want to find these differences between the two tables.
In what follows I will assume that if two columns, in the two tables, have the same name, they will also have the same data type. If that is not guaranteed in your case, it can be fixed with a little more work in the part where I identify the "common columns" - instead of matching them just by name, from the catalog views, they would have to be matched also by data type.
When you get to comparing rows in the two tables in the final step, (A minus B) union all (B minus A) works, but is not very efficient. Each table is read twice, and minus is an expensive operator. The more efficient solution, which I illustrate below, was discussed in a long thread on AskTom several years ago. Namely: collect all the rows from both tables (with union all), group by all the columns, and disregard the groups that have a count of 2. This means rows that were found in both tables, so they are duplicates in the union all! Actually, you will see a small additional trick to identify from which table the "non-duplicated" rows come. Add a column for "table_name" and in the final select, after grouping and keeping the groups with count(*) = 1, select max(table_name). You need an aggregate function (like max()) because you are grouping, but for these rows each group only has one row, so the max() is really just the table name.
The beauty of this approach is that it can be used to identify the common columns, too! In that case, we will compare rows from the USER_TAB_COLS view - we select column names that appear in either of the tables, and keep only the column names that are duplicates (so the column names appear in both tables). In that part of the solution, I also retrieve column_id, which is used to order the columns. Don't worry if you are not familiar with keep (dense_rank first...) - it's not really that complicated, but it's not that important either.
First let's set up a test case. I copy the EMP table from the SCOTT schema to my own schema, I replicate it (so now I have two copies, named EMP1 and EMP2), and I modify them slightly. I delete a different column from each, I delete a few (different) rows from each, and I modify one salary in one table. I will not show the resulting (slightly different) tables, but if you are following along, just select * from both and compare them before you continue reading.
Create the tables:
create table EMP1 as select * from scott.emp;
Table EMP1 created.
select * from EMP1;
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
----- ---------- --------- ---- ------------------- ----- ------ -------
7369 SMITH CLERK 7902 1980-12-17 00:00:00 800 20
7499 ALLEN SALESMAN 7698 1981-02-20 00:00:00 1600 300 30
7521 WARD SALESMAN 7698 1981-02-22 00:00:00 1250 500 30
7566 JONES MANAGER 7839 1981-04-02 00:00:00 2975 20
7654 MARTIN SALESMAN 7698 1981-09-28 00:00:00 1250 1400 30
7698 BLAKE MANAGER 7839 1981-05-01 00:00:00 2850 30
7782 CLARK MANAGER 7839 1981-06-09 00:00:00 2450 10
7788 SCOTT ANALYST 7566 1987-04-19 00:00:00 3000 20
7839 KING PRESIDENT 1981-11-17 00:00:00 5000 10
7844 TURNER SALESMAN 7698 1981-09-08 00:00:00 1500 0 30
7876 ADAMS CLERK 7788 1987-05-23 00:00:00 1100 20
7900 JAMES CLERK 7698 1981-12-03 00:00:00 950 30
7902 FORD ANALYST 7566 1981-12-03 00:00:00 3000 20
7934 MILLER CLERK 7782 1982-01-23 00:00:00 1300 10
Modify them slightly:
create table EMP2 as select * from EMP1;
Table EMP2 created.
alter table emp1 drop column hiredate;
Table EMP1 altered.
alter table emp2 drop column comm;
Table EMP2 altered.
delete from EMP1 where ename like 'A%';
2 rows deleted;
delete from EMP2 where sal >= 3000;
3 rows deleted
update EMP2 set sal = 2950 where empno = 7698;
1 row updated
commit;
At this point you would do well to select * from EMP1; and select * from EMP2; and compare.
Now let's find out what columns the two tables have left in common.
select column_name,
min(column_id) keep(dense_rank first order by table_name) as col_id
from user_tab_cols
where table_name in ('EMP1', 'EMP2')
group by column_name
having count(*) = 2
order by col_id;
COLUMN_NAME COL_ID
----------- ------
EMPNO 1
ENAME 2
JOB 3
MGR 4
SAL 5
DEPTNO 7
6 rows selected
Perfect, so now we can compare the two tables, but only after we "project" them along the common columns only.
select max(table_name) as table_name, EMPNO, ENAME, JOB, MGR, SAL, DEPTNO
from (
select 'EMP1' as table_name, EMPNO, ENAME, JOB, MGR, SAL, DEPTNO from EMP1
union all
select 'EMP2' as table_name, EMPNO, ENAME, JOB, MGR, SAL, DEPTNO from EMP2
)
group by EMPNO, ENAME, JOB, MGR, SAL, DEPTNO
having count(*) = 1
order by EMPNO, ENAME, JOB, MGR, SAL, DEPTNO, table_name;
TABLE_NAME EMPNO ENAME JOB MGR SAL DEPTNO
---------- ----- ---------- --------- ------ ------ --------
EMP2 7499 ALLEN SALESMAN 7698 1600 30
EMP1 7698 BLAKE MANAGER 7839 2850 30
EMP2 7698 BLAKE MANAGER 7839 2950 30
EMP1 7788 SCOTT ANALYST 7566 3000 20
EMP1 7839 KING PRESIDENT 5000 10
EMP2 7876 ADAMS CLERK 7788 1100 20
EMP1 7902 FORD ANALYST 7566 3000 20
7 rows selected
The output is pretty much what we needed. Notice the first column, which tells us where the "unpaired" row comes from; and note BLAKE, who has different salary in the two tables (and the first column helps us to see what salary he has in which table).
This looks perfect so far, but what to do when you have 1000 columns? You could put it together in C or Java etc., using the result from the "common columns" query above - or you could do it all in Oracle, with dynamic SQL.
As far as I know, there is no set limit on the length of the text of an SQL statement in Oracle; the documentation says "The limit on how long a SQL statement can be depends on many factors, including database configuration, disk space, and memory" (and probably on your Oracle version, which they didn't mention). In any case, it will be more than 4000 characters, so we need to work with CLOB. In particular, we can't use listagg() - we need a workaround. I use xmlagg() below. Then, the documentation says if you concatenate text and at least one operand is CLOB the result will be CLOB; if that doesn't work for you, you may have to wrap the smaller text fragments within to_clob(). The "dynamic SQL" query below will produce the full text of the query I used above; you will simply copy it and paste it back into your front-end and execute it. You may have to delete wrapping double-quotes or such, depending on your front-end and settings.
First here is how we can create a (potentially very long) string, the list of common column names, which is repeated five times in the final query - just look again at the "final query" we used to compare the two tables above.
with
common_cols ( column_name, col_id ) as (
select column_name,
min(column_id) keep(dense_rank first order by table_name) as col_id
from user_tab_cols
where table_name in ('EMP1', 'EMP2')
group by column_name
having count(*) = 2
),
col_string ( str ) as (
select rtrim(xmlcast(xmlagg(xmlelement(e, column_name, ', ') order by col_id)
as clob), ', ') from common_cols
)
select * from col_string;
STR
-----------------------------------
EMPNO, ENAME, JOB, MGR, SAL, DEPTNO
And finally the full dynamic SQL query (the result is exactly the query I used to compare EMP1 and EMP2 on their common columns earlier):
with
common_cols ( column_name, col_id ) as (
select column_name,
min(column_id) keep(dense_rank first order by table_name) as col_id
from user_tab_cols
where table_name in ('EMP1', 'EMP2')
group by column_name
having count(*) = 2
),
col_string ( str ) as (
select rtrim(xmlcast(xmlagg(xmlelement(e, column_name, ', ') order by col_id)
as clob), ', ') from common_cols
)
select 'select max(table_name) as table_name, ' || str || chr(10) ||
'from (' || chr(10) ||
' select ''EMP1'' as table_name, ' || str || ' from EMP1' || chr(10) ||
' union all' || chr(10) ||
' select ''EMP2'' as table_name, ' || str || ' from EMP2' || chr(10) ||
' )' || chr(10) ||
'group by ' || str || chr(10) ||
'having count(*) = 1' || chr(10) ||
'order by ' || str || ', table_name;' as comp_sql_str
from col_string;
I have an comma delimited string which I want to use in an "IN" clause of the statement.
eg: 100,101,102
Since In and "IN" clause I have to quote the individial strings, I use a replace function:
eg: select ''''||replace('100,101,102',',',''', ''')||'''' from dual;
The above query works, however, when I try to use the output of the above as an input to the "IN" clause, it returns no data. I am restricted by only SQL statements, so I cannot use PL/SQL code. Kindly help.
select * from employee where employee_number in (
select ''''||replace('100,101,102',',',''', ''')||'''' from dual);
The above does not work. Please let me know what I am missing.
The general approach in this case would be to parse the comma-separated list into an Oracle collection and to use that collection in your SQL statement. Tom Kyte has an example of this in his discussion on variable IN lists.
Assuming you create the myTableType type and the in_list function from that thread, you should be able to do
SELECT *
FROM employee
WHERE employee_number IN (
SELECT *
FROM TABLE( in_list( p_your_comma_separated_list ) )
)
pure SQL, but not very well tested...
select to_number(substr(postfix, 2, instr(postfix, ',' ,2)-2)) id
from (
select substr(val, instr(val, ',', 1, n)) postfix
from (select ',101,102,103,' val from dual)
, (
select level n
from dual
connect by level < 10)
where instr(val, ',', 1, n) > 0)
where instr(postfix, ',' ,2)> 2;
EDIT: improved
select substr(postfix, 1, instr(postfix, ',' ,1)-1)
from (
select substr(val, instr(val, ',',1, level)+1) postfix
from (select ',101,102,103,' val from dual)
connect by instr(val, ',', 2, level) > 0
);
Note:
pre/post fix your strings with comma
adopt the upper limit (10 in the example) as per your needs (not needed in the improved version).
use the in_list table function mentioned by Justing Cave, that's probably better :)
credit: something like that is in Stephane Faroult's book "Refactorying SQL Applications" (O'Reilly)
As the comma-separated values contain only digits, why not try something as simple as using:
INSTR(','||my_csv_list_of_values||',', ','||my_search_value||',') <> 0
See this example:
-- some test data
with employee as (
select 101 as employee_number from dual
union select 200 from dual
union select 10 from dual
union select 102 from dual)
-- the actual query
select * from employee
where INSTR(','||'101,102,103,104'||',', ','||employee_number||',') <> 0;
-- ^^^^^^^^^^^^^^^^^
-- your CSV data
Producing:
EMPLOYEE_NUMBER
101
102
You can use your approach with REPLACE and IN if you format the entire select as a string - then use the string with either OPEN refcursor FOR or EXECUTE IMMEDIATE.
You can use XMLTABLE to convert comma separated string to a table.
Example:
select * from employee where employee_number in (
SELECT TO_NUMBER (column_value)
FROM XMLTABLE('100,101,102') );
You can use regexp_substr function to get expected output. For Example
NAMES:='SMITH,ALLEN,WARD,JONES'; -- here "NAMES" is the variable/result of expected input.that can be use in IN Clause.
SQL> select regexp_substr(NAMES,'[^,]+', 1, level) from dual 2 connect by regexp_substr(NAMES, '[^,]+', 1, level) is not null;
REGEXP_SUBSTR('SMITH,A
----------------------
SMITH
ALLEN
WARD
JONES
The above query iterates through the comma separated string, searches for the comma (,) and then splits the string by treating the comma as delimiter. It returns the string as a row, whenever it hits a delimiter.
Here is an Reference Click Here
SQL> select * from emp where ename in (
2 select regexp_substr('SMITH,ALLEN,WARD,JONES','[^,]+', 1, level) from dual
3 connect by regexp_substr('SMITH,ALLEN,WARD,JONES', '[^,]+', 1, level) is not null );
EMPNO ENAME JOB MGR HIREDATE SAL COMM DEPTNO
---------- ---------- --------- ---------- --------- ---------- ---------- ----------
7369 SMITH CLERK 7902 17-DEC-80 800 20
7499 ALLEN SALESMAN 7698 20-FEB-81 1600 300 30
7521 WARD SALESMAN 7698 22-FEB-81 1250 500 30
7566 JONES MANAGER 7839 02-APR-81 2975 20